N Queens Problem

The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, following is a solution for 4 Queen problem.

The expected output is a binary matrix which has 1s for the blocks where queens are placed. For example following is the output matrix for above 4 queen solution.

{ 0, 1, 0, 0}
{ 0, 0, 0, 1}
{ 1, 0, 0, 0}
{ 0, 0, 1, 0}

Naive Algorithm
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.

while there are untried conflagrations
{
generate the next configuration
if queens don't attack in this configuration then
{
print this configuration;
}
}

Backtracking Algorithm
The idea is to place queens one by one in different columns, starting from the leftmost column. When we place a queen in a column, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes then we backtrack and return false.

1) Start in the leftmost column
2) If all queens are placed
return true
3) Try all rows in the current column. Do following for every tried row.
a) If the queen can be placed safely in this row then mark this [row,
column] as part of the solution and recursively check if placing
queen here leads to a solution.
b) If placing queen in [row, column] leads to a solution then return
true.
c) If placing queen doesn't lead to a solution then umark this [row,
column] (Backtrack) and go to step (a) to try other rows.
3) If all rows have been tried and nothing worked, return false to trigger
backtracking.

Implementation of Backtracking solution

#define N 4

#include<stdio.h>

/* A utility function to print solution */

voidprintSolution(intboard[N][N])

{

for(inti = 0; i < N; i++)

{

for(intj = 0; j < N; j++)

printf(" %d ", board[i][j]);

printf("n");

}

}

/* A utility function to check if a queen can be placed on board[row][col]

Note that this function is called when "col" queens are already placeed

in columns from 0 to col -1. So we need to check only left side for

attacking queens */

boolisSafe(intboard[N][N], introw, intcol)

{

inti, j;

/* Check this row on left side */

for(i = 0; i < col; i++)

{

if(board[row][i])

returnfalse;

}

/* Check upper diagonal on left side */

for(i = row, j = col; i >= 0 && j >= 0; i--, j--)

{

if(board[i][j])

returnfalse;

}

/* Check lower diagonal on left side */

for(i = row, j = col; j >= 0 && i < N; i++, j--)

{

if(board[i][j])

returnfalse;

}

returntrue;

}

/* A recursive utility function to solve N Queen problem */

boolsolveNQUtil(intboard[N][N], intcol)

{

/* base case: If all queens are placed then return true */

if(col >= N)

returntrue;

/* Consider this column and try placing this queen in all rows

one by one */

for(inti = 0; i < N; i++)

{

/* Check if queen can be placed on board[i][col] */

if( isSafe(board, i, col) )

{

/* Place this queen in board[i][col] */

board[i][col] = 1;

/* recur to place rest of the queens */

if( solveNQUtil(board, col + 1) == true)

returntrue;

/* If placing queen in board[i][col] doesn't lead to a solution

then remove queen from board[i][col] */

board[i][col] = 0; // BACKTRACK

}

}

/* If queen can not be place in any row in this colum col

then return false */

returnfalse;

}

/* This function solves the N Queen problem using Backtracking. It mainly uses

solveNQUtil() to solve the problem. It returns false if queens cannot be placed,

otherwise return true and prints placement of queens in the form of 1s. Please

note that there may be more than one solutions, this function prints one of the